: +5V

: ,----------,

: | |

: | |

: | \

: | / R1

: | \ ???

: | /

: | |

: --- | +2V

: - V1 +--------,

: --- 5V | |

: - | |

: | \ \

: | / R2 / Rload

: | \ ??? \ 2V @ 250mA

: | / /

: | | |

: | | |

: '----------+--------'

Suppose you transform this, using an NPN transistor:

: +5V

: ,----------+-------,

: | | |

: | | | almost 250mA

: | \ | |

: | / R1 | |

: | \ ??? | v

: | / |

: | | |

: --- | |/ Q1

: - V1 +-----| NPN

: --- 5V | |>-,

: - | | <---- supposed to be about 2V

: | \ \

: | / R2 / Rload

: | \ ??? \ 2V @ 250mA

: | / /

: | | |

: | | |

: '----------+--------'

Now, in this case you can use far less current in your R1,R2 divider

because the transistor's base will not divert much current away from

it.

Q1 will bring pretty much all of the 250mA used in Rload via the

collector that connects directly back to the +5 side of the 5V battery

or power supply. As a 0-order estimate of Q1's base voltage, you can

just assume that the base-to-emitter voltage (Vbe) is somewhere

between 0.6V (lower Vbe voltages occur with smaller base currents,

larger Vbe with larger base currents, and it is unusual for 0.6V to be

the case unless we are talking about fairly low currents overall) to

about 0.9V or slightly more (for times when the transistor is involved

with fairly high currents.)

Normally, this variation of Vbe is about 60mV for each factor of 10

change in current, so you can see that it varies only slightly. So

let's say it runs at about 750mV with a base current of 100uA, just to

pick some figure to start. If the base current went to 1mA, you'd

expect the Vbe voltage to then be about 750mV + 60mV or about 810mV.

That should give you a rough picture, here.

Getting back to the diagram... Rload needs about 250mA per your

specification. This means a collector current of about the same

value. But what base current? (We need to have a rough idea of the

base current to estimate Vbe -- not that it is terribly important to

do, as it isn't because of the log() behavior, but let's do it

anyway.)

Well, if the transistor is to be operated in saturation that would be

one thing. But it isn't in this case. The collector voltage is known

to be 5V and the emitter is at 2V, roughly. What causes a transistor

to go into saturation, so to speak, is when the voltage potential at

the collector is such that it begins to even slightly forward bias the

base to collector diode in it (it's supposed to be normally reverse

biased.) In this case, we already know that the base won't be more

than about 3V (we are still trying to figure exactly what, but we can

delay this for now) and with the collector at 5V the internal base to

collector diode will definitely be reverse biased. So no saturation

to worry about.

Okay, so that means that we can anticipate a transistor beta of say

100 or more. Very old transistors, like point-contact types, might

have betas in the 25-30 range. But none of those I've used recently

are that "bad" unless they are in reasonable saturation (when the

forward biased base-to-collector diode demolishes any decent beta

computation.) You could probably even be safe thinking 200. But

let's call it 100 to be extra safe in the other figures we derive.

So the base current will be about 1/100th of the 250mA. This means...

2.5mA. Going back to the 100uA starting point I mentioned, this is a

factor of 25 times more or about a fourth of 10x10. The first 10x

gets us another 60mV. Let's not get into log() functions and just

accept that the next 2.5 times will about to 1/4th of the 60mV more,

or 15mV. So that means 75mV or so above our guess of 750mV for the

NPN, or 825mV. Call it 850mV.

Okay, so you want the base to be about 850mV above 2.0V or 2.85V. But

notice now that your load on the R1,R2 divider has been reduced to the

base current which we estimate at 2.5mA -- maybe even less than that.

This means that all of the same calculations performed before apply

here again -- except that you now what the divider to provide 2.85V

instead of the original 2V or so and that you now only need to supply

2.5mA or so instead of 250mA.

With no load at all, or better for thinking about, close to no load

but with a very slight load, the Vbe will be around 0.6V or so. This

means that if we want there to be 2.2V unloaded and 1.9V loaded down,

we want the base to be 2.2+0.6 or 2.8V unloaded and then about 1.9+.85

or 2.75V loaded. A change of 50mV total. Remember this figure, as

I'll use it shortly in (2) below.

Referring back to the old equations and the Thevenin equivalent,

updated now with new values, we have:

(1) 5*R2/(R1+R2) = 2.8 --or-- R2/(R1+R2) = 2.8/5

--and also that--

(2) Rth=R1*R2/(R1+R2) = dV/dI = (50mV/2.5mA) = 20

The 50mV comes from the voltage change noted above. The 2.5mA comes

from the change in base current between an unloaded case and a loaded

one.

This solves out to R1 = 20*5/2.8 = 100/2.8 = 35.7 ohms and to R2 =

R1*(2.8/5)/(1-2.8/5) = 45 4/9 = 45.5 ohms.

Now we are talking about only 60mA or so through the divider with

250mA through Rload and the wattage in the divider is about 300mW

instead of 5W.

There will be another (5V - 1.9V)*250mA or 775mW in the transistor, as

well. But that waste is unavoidable in linear (non-switching) designs

because you simply _have_ to throw away about 3V at 250mA no matter

what you do. So that 3/4 watt is going to be wasted somewhere.

So this design is about 300mW more than the theoretical best case for

a linear arrangement (which must be 750mW, as I argue) rather than

being more than 4W beyond, like the simple divider was. And all you

had to do was add a transistor to get there and redo a few

calculations for the resistors.

You could plan on further reductions on the base current by adding

another transistor. But all you could hope to do is eat into that

300mW and reduce it somewhat. But that 750mW waste will always remain

no matter how many transistors you stack up, unless you consider a

switcher arrangement.

Jon